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प्रश्न
Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 μF. When the ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network.
उत्तर
The equivalent circuit is given below.
There are two capacitors in one branch in series. So, the equivalent capacitance of one branch will be
1/1+1/1=2muF
The arrangement will be further reduced to the form given below.
Now, both the capacitors are in parallel, so the equivalent capacitance will be
2+2=4μF
Therefore, the equivalent capacitance is 4 μF.
(i)
Voltage, V = 6 V
The charge in the network is given by
q=CV
Here, C is the equivalent capacitance.
Now,q=4×10−6×6
=24×10−6 C=24 μC
(ii)
The energy stored in the network is given by
E=1/2CV2
=12×4×10−6×(6)2
=12×4×36×10−6
=72×10−6 J
=72 μJ
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